A calorimeter of water equivalent 20 g contains 180g of water at 250C. ‘m’ grams of steam at 1000C is mixed in it till the temperature of the mixture is 310C. The value of ‘m’ is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1 0C-1)

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3.2

2.6

answer is A.

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Detailed Solution

Heat gained by cold body = heat lost by hot body⇒20+180×1×31-25=m(540)+m×1×100-31⇒1200=m(609)⇒m≈2gms

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